The Stacks project

Proof. Observe that $A^\wedge$ is reduced, see More on Algebra, Lemma 15.43.6. The rings $A^ h$ and $A^{sh}$ are reduced by More on Algebra, Lemma 15.45.4. The dimensions of $A$, $A^ h$, $A^{sh}$, and $A^\wedge$ are the same by More on Algebra, Lemmas 15.43.1 and 15.45.7.

Recall that a Noetherian local ring is Nagata if and only if the formal fibres of $A$ are geometrically reduced, see More on Algebra, Lemma 15.52.4. This property is inherited by $A^ h$ and $A^{sh}$, see the material in More on Algebra, Section 15.51 and especially Lemma 15.51.8. The completion is Nagata by Algebra, Lemma 10.162.8.

Now we come to the statement on integral closures. Before continuing let us pick $f \in \mathfrak m$ as in Lemma 33.39.1. Then the image of $f$ in $A^ h$, $A^{sh}$, and $A^\wedge$ clearly is an element satisfying properties (1) – (6) in that ring.

Since $A \to A'$ is finite we see that $A' \otimes _ A A^ h$ and $A' \otimes _ A A^{sh}$ is the product of henselian local rings finite over $A^ h$ and $A^{sh}$, see Algebra, Lemma 10.153.4. Each of these local rings is the henselization of $A'$ at a maximal ideal $\mathfrak m' \subset A'$ lying over $\mathfrak m$, see Algebra, Lemma 10.156.1 or 10.156.3. Hence these local rings are normal domains by More on Algebra, Lemma 15.45.6. It follows that $A' \otimes _ A A^ h$ and $A' \otimes _ A A^{sh}$ are normal rings. Since $A^ h \to A' \otimes _ A A^ h$ and $A^{sh} \to A' \otimes _ A A^{sh}$ are finite (hence integral) and since $A' \otimes _ A A^ h \subset (A^ h)_ f = Q(A^ h)$ and $A' \otimes _ A A^{sh} \subset (A^{sh})_ f = Q(A^{sh})$ we conclude that $A' \otimes _ A A^ h$ and $A' \otimes _ A A^{sh}$ are the desired integral closures.

For the completion we argue in entirely the same manner. First, by Algebra, Lemma 10.97.8 we have

The local rings $A'_{\mathfrak m'}$ are normal and have dimension $1$ (by Algebra, Lemma 10.113.2 for example or the discussion in Algebra, Section 10.112). Thus $A'_{\mathfrak m'}$ is a discrete valuation ring, see Algebra, Lemma 10.119.7. Hence $(A'_{\mathfrak m'})^\wedge$ is a discrete valuation ring by More on Algebra, Lemma 15.43.5. It follows that $A' \otimes _ A A^\wedge$ is a normal ring and we can conclude in exactly the same manner as before. $\square$